3.1066 \(\int (1-x)^{3/2} \sqrt{1+x} \, dx\)

Optimal. Leaf size=48 \[ \frac{1}{3} (1-x)^{3/2} (x+1)^{3/2}+\frac{1}{2} \sqrt{1-x} x \sqrt{x+1}+\frac{1}{2} \sin ^{-1}(x) \]

[Out]

(Sqrt[1 - x]*x*Sqrt[1 + x])/2 + ((1 - x)^(3/2)*(1 + x)^(3/2))/3 + ArcSin[x]/2

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Rubi [A]  time = 0.0056664, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {49, 38, 41, 216} \[ \frac{1}{3} (1-x)^{3/2} (x+1)^{3/2}+\frac{1}{2} \sqrt{1-x} x \sqrt{x+1}+\frac{1}{2} \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(1 - x)^(3/2)*Sqrt[1 + x],x]

[Out]

(Sqrt[1 - x]*x*Sqrt[1 + x])/2 + ((1 - x)^(3/2)*(1 + x)^(3/2))/3 + ArcSin[x]/2

Rule 49

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*(m
+ n + 1)), x] + Dist[(2*c*n)/(m + n + 1), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x]
&& EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0] && IGtQ[n + 1/2, 0] && LtQ[m, n]

Rule 38

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(x*(a + b*x)^m*(c + d*x)^m)/(2*m + 1)
, x] + Dist[(2*a*c*m)/(2*m + 1), Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0]

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int (1-x)^{3/2} \sqrt{1+x} \, dx &=\frac{1}{3} (1-x)^{3/2} (1+x)^{3/2}+\int \sqrt{1-x} \sqrt{1+x} \, dx\\ &=\frac{1}{2} \sqrt{1-x} x \sqrt{1+x}+\frac{1}{3} (1-x)^{3/2} (1+x)^{3/2}+\frac{1}{2} \int \frac{1}{\sqrt{1-x} \sqrt{1+x}} \, dx\\ &=\frac{1}{2} \sqrt{1-x} x \sqrt{1+x}+\frac{1}{3} (1-x)^{3/2} (1+x)^{3/2}+\frac{1}{2} \int \frac{1}{\sqrt{1-x^2}} \, dx\\ &=\frac{1}{2} \sqrt{1-x} x \sqrt{1+x}+\frac{1}{3} (1-x)^{3/2} (1+x)^{3/2}+\frac{1}{2} \sin ^{-1}(x)\\ \end{align*}

Mathematica [A]  time = 0.0375313, size = 44, normalized size = 0.92 \[ \frac{1}{6} \left (-2 x^2+3 x+2\right ) \sqrt{1-x^2}-\sin ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - x)^(3/2)*Sqrt[1 + x],x]

[Out]

((2 + 3*x - 2*x^2)*Sqrt[1 - x^2])/6 - ArcSin[Sqrt[1 - x]/Sqrt[2]]

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Maple [B]  time = 0.003, size = 71, normalized size = 1.5 \begin{align*}{\frac{1}{3} \left ( 1-x \right ) ^{{\frac{3}{2}}} \left ( 1+x \right ) ^{{\frac{3}{2}}}}+{\frac{1}{2}\sqrt{1-x} \left ( 1+x \right ) ^{{\frac{3}{2}}}}-{\frac{1}{2}\sqrt{1-x}\sqrt{1+x}}+{\frac{\arcsin \left ( x \right ) }{2}\sqrt{ \left ( 1+x \right ) \left ( 1-x \right ) }{\frac{1}{\sqrt{1-x}}}{\frac{1}{\sqrt{1+x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-x)^(3/2)*(1+x)^(1/2),x)

[Out]

1/3*(1-x)^(3/2)*(1+x)^(3/2)+1/2*(1-x)^(1/2)*(1+x)^(3/2)-1/2*(1-x)^(1/2)*(1+x)^(1/2)+1/2*((1+x)*(1-x))^(1/2)/(1
+x)^(1/2)/(1-x)^(1/2)*arcsin(x)

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Maxima [A]  time = 1.55182, size = 38, normalized size = 0.79 \begin{align*} \frac{1}{3} \,{\left (-x^{2} + 1\right )}^{\frac{3}{2}} + \frac{1}{2} \, \sqrt{-x^{2} + 1} x + \frac{1}{2} \, \arcsin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)^(3/2)*(1+x)^(1/2),x, algorithm="maxima")

[Out]

1/3*(-x^2 + 1)^(3/2) + 1/2*sqrt(-x^2 + 1)*x + 1/2*arcsin(x)

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Fricas [A]  time = 1.53546, size = 124, normalized size = 2.58 \begin{align*} -\frac{1}{6} \,{\left (2 \, x^{2} - 3 \, x - 2\right )} \sqrt{x + 1} \sqrt{-x + 1} - \arctan \left (\frac{\sqrt{x + 1} \sqrt{-x + 1} - 1}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)^(3/2)*(1+x)^(1/2),x, algorithm="fricas")

[Out]

-1/6*(2*x^2 - 3*x - 2)*sqrt(x + 1)*sqrt(-x + 1) - arctan((sqrt(x + 1)*sqrt(-x + 1) - 1)/x)

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Sympy [B]  time = 4.92475, size = 168, normalized size = 3.5 \begin{align*} \begin{cases} - i \operatorname{acosh}{\left (\frac{\sqrt{2} \sqrt{x + 1}}{2} \right )} - \frac{i \left (x + 1\right )^{\frac{7}{2}}}{3 \sqrt{x - 1}} + \frac{11 i \left (x + 1\right )^{\frac{5}{2}}}{6 \sqrt{x - 1}} - \frac{17 i \left (x + 1\right )^{\frac{3}{2}}}{6 \sqrt{x - 1}} + \frac{i \sqrt{x + 1}}{\sqrt{x - 1}} & \text{for}\: \frac{\left |{x + 1}\right |}{2} > 1 \\\operatorname{asin}{\left (\frac{\sqrt{2} \sqrt{x + 1}}{2} \right )} + \frac{\left (x + 1\right )^{\frac{7}{2}}}{3 \sqrt{1 - x}} - \frac{11 \left (x + 1\right )^{\frac{5}{2}}}{6 \sqrt{1 - x}} + \frac{17 \left (x + 1\right )^{\frac{3}{2}}}{6 \sqrt{1 - x}} - \frac{\sqrt{x + 1}}{\sqrt{1 - x}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)**(3/2)*(1+x)**(1/2),x)

[Out]

Piecewise((-I*acosh(sqrt(2)*sqrt(x + 1)/2) - I*(x + 1)**(7/2)/(3*sqrt(x - 1)) + 11*I*(x + 1)**(5/2)/(6*sqrt(x
- 1)) - 17*I*(x + 1)**(3/2)/(6*sqrt(x - 1)) + I*sqrt(x + 1)/sqrt(x - 1), Abs(x + 1)/2 > 1), (asin(sqrt(2)*sqrt
(x + 1)/2) + (x + 1)**(7/2)/(3*sqrt(1 - x)) - 11*(x + 1)**(5/2)/(6*sqrt(1 - x)) + 17*(x + 1)**(3/2)/(6*sqrt(1
- x)) - sqrt(x + 1)/sqrt(1 - x), True))

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Giac [A]  time = 1.0681, size = 59, normalized size = 1.23 \begin{align*} -\frac{1}{3} \,{\left (x + 1\right )}^{\frac{3}{2}}{\left (x - 1\right )} \sqrt{-x + 1} + \frac{1}{2} \, \sqrt{x + 1} x \sqrt{-x + 1} + \arcsin \left (\frac{1}{2} \, \sqrt{2} \sqrt{x + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)^(3/2)*(1+x)^(1/2),x, algorithm="giac")

[Out]

-1/3*(x + 1)^(3/2)*(x - 1)*sqrt(-x + 1) + 1/2*sqrt(x + 1)*x*sqrt(-x + 1) + arcsin(1/2*sqrt(2)*sqrt(x + 1))